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Notes from Lecture 4: Ultrafilter Extensions and the Standard Translation Eric Pacuit∗ March 1, 2012 1 Ultrafilter Extensions Definition 1.1 (Ultrafilter) Let W be a non-empty set. An ultrafilter on W is a set u ⊆ }(W ) satisfying the following properties: 1. ; 62 u 2. If X; Y 2 u then X \ Y 2 u 3. If X 2 u and X ⊆ Y then Y 2 u. 4. For all X ⊆ W , either X 2 u or X 2 u (where X is the complement of X in W ) / A collection u0 ⊆ }(W ) has the finite intersection property provided for each X; Y 2 u0, X \ Y 6= ;. Theorem 1.2 (Ultrafilter Theorem) If a set u0 ⊆ }(W ) has the finite in- tersection property, then u0 can be extended to an ultrafilter over W (i.e., there is an ultrafilter u over W such that u0 ⊆ u. Proof. Suppose that u0 has the finite intersection property. Then, consider the set u1 = fZ j there are finitely many sets X1;:::Xk such that Z = X1 \···\ Xkg: That is, u1 is the set of finite intersections of sets from u0. Note that u0 ⊆ u1, since u0 has the finite intersection property, we have ; 62 u1, and by definition u1 is closed under finite intersections. Now, define u2 as follows: 0 u = fY j there is a Z 2 u1 such that Z ⊆ Y g ∗UMD, Philosophy. Webpage: ai.stanford.edu/∼epacuit, Email: [email protected] 1 0 0 We claim that u is a consistent filter: Y1;Y2 2 u then there is a Z1 2 u1 such that Z1 ⊆ Y1 and Z2 2 u1 such that Z2 ⊆ Y2. Then, since ; 6= Z1 \ Z2 2 u1, we have Z1 \ Z2 ⊆ Y1 \ Y2. Hence, Y1 \ Y2 2 u1. Also, if X 2 u1 then there is a Z 2 u1 such that Z ⊆ X. If X ⊆ Y , then Z ⊆ Y and so Y 2 u1. Hence, u1 is a consistent filter. The next step is to show that u1 can be extended to an ultrafilter. This follows almost directly from Zorn's Lemma1: Consider the set Z of all filters that extend u1. That is, Z = fv j u1 ⊆ v and v is a consistent filterg. Note that Z is partially-ordered under the ⊆-relation. Furthermore, the upper bound of any chain in Z (i.e., sequence of ultrafilters v0 ⊆ v1 ⊆ · · · ) is the union of all the filters in the chain. This collection of sets will be a consistent ultrafilter extending u1, and so is contained in Z. By Zorn's Lemma, Z must contain a maximal element. This maximal element must be an ultrafilter (containing u1). qed Let M = hW; R; V i be a Kripke model. Two functions are relevant to our analysis: • m : }(W ) ! }(W ) defined as m(X) = fw j there is a v such that wRv and v 2 Xg, and • l : }(W ) ! }(W ) defined as l(X) = fw j for all v, if wRv then v 2 Xg. Definition 1.3 (Ultrafilter Extension) An ultrafilter extension is a model ue(M) = hUf(W );Rue;V uei where • Uf(W ) = fu j u is an ultrafilter over W g, • uRueu0 iff for all X ⊆ W , if X 2 u0 then m(X) 2 u, and • V ue(p) = fu j V (p) 2 ug. / Fact 1.4 In an ultrafilter extension ue(M) = hUf(W );Rue;V uei, we have uRueu0 iff fY j l(Y ) 2 ug ⊆ u0. Proof. Left as an exercise. qed Let M = hW; R; V i be a model . The truth map [[·]]M : L! }(W ) is defined by induction on the structure of ' as follows: • For p 2 At, [[p]]M = V (p) 1Zorn's Lemma states that in a partially ordered set P , if every chain has an upper bound in P , then P contains at least on maximal element. The proof of Zorn's Lemma uses the Axiom of Choice (indeed, it is equivalent to the Axiom of Choice). 2 • [[:']]M = W − [[']]M • [[' ^ ]]M = [[']]M \ [[ ]]M • [[3']]M = m([[']]M) Lemma 1.5 For all models M = hW; R; V i, for all modal formulas ', we have [[']]M 2 u iff ue(M); u j= '. Proof. The proof is by induction on the structure of '. Then we have, Base case: ' is p 2 At. [[p]]M 2 u iff V (p) 2 u (Definition of [[·]]M) iff u 2 V ue(p) (Definition of V ue) iff ue(M); u j= p (Definition of truth in a model) 2 Induction Hypothesis: For all less complex than ', [[ ]]M 2 u iff ue(M); u j= Case 1: ' is : : [[: ]]M 2 u iff W − [[ ]]M 2 u (Definition of [[·]]M) iff [[ ]]M 62 u (Properties of an ultrafilter) iff ue(M); u 6j= (Induction Hypothesis) iff ue(M); u j= : (Definition of truth) Case 2: ' is 1 ^ 2 [[ 1 ^ 2]]M 2 u iff [[ 1]]M \ [[ 2]]M 2 u (Definition of [[·]]M) iff [[ 1]]M 2 u and [[ 2]]M 2 u (Properties of ultrafilters) iff ue(M); u j= 1 and ue(M); u j= 2 (Induction hypothesis) iff ue(M); u j= 1 ^ 2 (Definition of truth) Case 3: ' is 3 Suppose that ue(M); u j= 3 . Then, there is a u0 2 Uf(W ) such that uRueu0 0 0 and ue(M); u j= . By the induction hypothesis, [[ ]]M 2 u . By the definition ue of R , we have m([[ ]]M) 2 u, and so, [[3 ]]M 2 u. Thus, we have shown that ue(M); u j= 3 implies [[3 ]]M 2 u. Suppose that [[3 ]]M 2 u. We must show ue(M); u j= 3 . Consider the set u0 = fY j l(Y ) 2 ug [ f[[ ]]Mg 2Less complex means that contains fewer connectives. 3 We claim that u0 has the finite intersection property. We first show that fY j l(Y ) 2 ug is closed under finite intersections. It is enough to show that for any two sets Y1;Y2 2 fY j l(Y ) 2 ug, Y1 \ Y2 2 fY j l(Y ) 2 ug (why?). Suppose that Y1;Y2 2 fY j l(Y ) 2 ug. Note that l(Y1 \ Y2) = l(Y1) \ l(Y2). Then, since u is an ultrafilter and l(Y1); l(Y2) 2 u, we have l(Y1 \ Y2) = l(Y1) \ l(Y2) 2 u. Hence Y1 \ Y2 2 fY j l(Y ) 2 ug. Next we show that for any Z 2 fY j l(Y ) 2 ug, we have Z \ [[ ]]M 6= ;. Choose an arbitrary Z such that l(Z) 2 u. We will show Z \ [[ ]]M 6= ;. Since l(Z) 2 u and [[3 ]]M 2 u, we have l(Z) \ [[3 ]]M 2 u, and so l(Z) \ [[3 ]]M 6= ;. Let w 2 l(Z) \ [[3 ]]M. Then, there is a v 2 W such that M; v j= . I.e., wRv and v 2 [[ ]]M. Since w 2 l(Y ) and wRv, we have v 2 Y . Hence, v 2 Y \ [[ ]]M. This implies that u0 has the finite intersection property (why?). 0 0 By the ultrafilter theorem, there is an ultrafilter u such that u0 ⊆ u . Since fY j l(Y ) 2 ug ⊆ u0, we have uRueu0. By the induction hypothesis, since 0 0 [[ ]]M 2 u , we have ue(M); u j= . Hence, ue(M); u j= 3 . qed Corollary 1.6 For all models M and states w in M, we have w ! uw, where uw is the principle ultrafilter generated by w. Proof. Let M = hW; R; V i be a model and w 2 W . The principle ultrafilter generated by w is uw = fX ⊆ W j w 2 Xg. Let ' be an arbitrary modal formula. We have M; w j= ' iff w 2 [[']]M iff [[']]M 2 uw iff ue(M); uw j= ' (the latter equivalence follows from the above Lemma). qed Lemma 1.7 For all models M, ue(M) is m-saturated. Proof. Suppose that ue(M) = hUf(W );Rue;V uei an ultrafilter extension of some model M = hW; R; V i. Let u 2 Uf(W ) be any state in ue(M) and Σ be a arbitrary set of modal formulas. Suppose that every finite subset of Σ is satisfiable at some successor of u (i.e., for each finite set ∆ ⊆ Σ, there is a ue V state v∆ 2 W such that uR v∆ and ue(M); v∆ j= ∆). We must find a state v 2 W such that uRuev and ue(M); v j= Σ (i.e., for each 2 Σ, ue(M); v j= ). Consider the set v0 = fY j l(Y ) 2 ug [ f[[ ]]M j 2 Σg We will show v0 has the finite intersection property. Since fY j l(Y ) 2 ug is T closed under finite intersections, it is enough to show that Y \ f[[ ]]M j 2 T V ∆g 6= ; for some finite subset ∆ of Σ. Note that f[[ ]]M j 2 ∆g = [[ ∆]]M. ue Recall that ∆ is satisfiable at some successor state v∆ of u. That is, uR v∆ V V and ue(M); v∆ j= ∆. By Lemma 1.5, this means [[ ∆]]M 2 v∆. By the ue V V definition of R , we have m([[ ∆]]M) 2 u. Hence, [[3 ∆]]M 2 u. Since u is 4 V an ultrafilter, we have l(Y ) \ [[3 ∆]]M 2 u. Hence, (since ; 62 u) there is a V V x 2 l(Y ) \ [[3 ∆]]M. This implies there is a y such that xRy and y 2 [[ ∆]]M.
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